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Help me to work out a problem: a thing has nine sides, the probability of each side is one ninth, continuous throwing 7 times, at least once, what is the probability of the ninth side
The probability of each face is 1 / 9, so the probability of the ninth face: 1 / 9
Probability of the ninth side not appearing: 1-1 / 9 = 8 / 9
The probability of no ninth face in 7 consecutive tosses: (8 / 9) ^ 7 = 0.4384
So the probability of the ninth plane appearing at least once: 1-0.4384 = 0.5616
1 of 3 * 5 + 1 of 5 * 7 + 1 of 7 * 9 + 1 of 9 * 11 + 1 of 11 * 13
The original formula = 1 / 2x (1 / 3-1 / 5) + 1 / 2x (1 / 5-1 / 7) + 1 / 2x (1 / 7-1 / 9) + 1 / 2x (1 / 9-1 / 11) + 1 / 2x (1 / 11-1 / 13) = 1 / 2x (1 / 3-1 / 5 + 1 / 5-1 / 7 + 1 / 7-1 / 9 + 1 / 9-1 / 11 + 1 / 11-1 / 13) = 1 / 2x (1 / 3-1 / 13) = 1 / 2x (13 / 39-3 / 39) = 5 / 39 & nbsp; (5 / 39)
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